1. The radioisotope Cobalt-60 is used in cancer therapy. The half-life of this isotope is 5.27 years.

Which equation determines the percent of an initial amount of the isotope remaining after t years?

A=5.27(12)100tA=5.27(12)100t

A=100(12)t5.27A=100(12)t5.27

A=5.27(12)t100A=5.27(12)t100

A=100(12)5.27t

**My Answer-'C'**

2. What are the real and imaginary parts of the complex number?

9+7i

The real part:

The imaginary part:

**My Answer- Real part=9, Imag. part=7i**

3.

What is the simplest radical form of the expression?

(x4y7)23

xy2xy−−√3xy2xy3

x6y10y√x6y10y

x2y4x2y2−−−−√3x2y4x2y23

x3y5x√

**My Answer-'D'**

**Thanks so much**

Guest Jan 16, 2018

#1**+1 **

The correct equation in the first part should be

A = 100(1/2)^{t/5.27}

To see that this is correct....let t = 5.27 and we have that

A = 100(1/2)^{ 5.27/5.27} = 100 (1/2)^{1} = 100 (1/2) = 50% left after 5.27 years, the half-life

Your second answer is correct

I can't tell exactly what the third one is....sorry....!!!

CPhill
Jan 16, 2018